If y = cos^{-1}(tanh x) + sinh(sin 6x),then f | vedclass

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Question

(A) Given $y = \cos^{-1}(	anh x) + \sinh(\sin 6x)$.Applying the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\cos^{-

Vedclass · Accepted Answer

$\frac{-1}{\cosh x} + 6 \cos 6x \cosh(\sin 6x)$

Vedclass · Answer

(A) Given $y = \cos^{-1}(	anh x) + \sinh(\sin 6x)$.Applying the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\cos^{-1}(	anh x)) + \frac{d}{dx}(\sinh(\sin 6x))$$\frac{dy}{dx} = \frac{-1}{\sqrt{1 - 	anh^2 x}} \cdot \frac{d}{dx}(	anh x) + \cosh(\sin 6x) \cdot \frac{d}{dx}(\sin 6x)$Using the identities $1 - 	anh^2 x = 	ext{sech}^2 x$ and $\frac{d}{dx}(	anh x) = 	ext{sech}^2 x$:$\frac{dy}{dx} = \frac{-1}{\sqrt{	ext{sech}^2 x}} \cdot 	ext{sech}^2 x + \cosh(\sin 6x) \cdot (6 \cos 6x)$$\frac{dy}{dx} = \frac{-1}{	ext{sech} x} \cdot 	ext{sech}^2 x + 6 \cos 6x \cosh(\sin 6x)$$\frac{dy}{dx} = -	ext{sech} x \cdot 	ext{sech} x \cdot \cosh x + 6 \cos 6x \cosh(\sin 6x)$ (Note: $	ext{sech} x = \frac{1}{\cosh x}$)$\frac{dy}{dx} = \frac{-1}{\cosh x} + 6 \cos 6x \cosh(\sin 6x)$.Thus,the correct option is $A$.

If $y = \cos^{-1}(\tanh x) + \sinh(\sin 6x)$,then $\frac{dy}{dx} =$

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